Operations on Whole Numbers - Exercise 4.5 Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : Operations on Whole Numbers - Exercise 4.5 Class 6 Notes | EduRev

 Page 1


 
 
 
 
 
 
Exercise 4.5                                                                           page: 4.23 
1. Without drawing a diagram, find 
(i) 10
th
 square number 
(ii) 6
th
 triangular number 
Solution: 
 
(i) 10
th
 square number 
The square number can be remembered using the following rule 
Nth square number = n × n 
So the 10
th
 square number = 10 × 10 = 100 
 
(ii) 6
th
 triangular number 
The triangular number can be remembered using the following rule 
Nth triangular number = n × (n + 1)/ 2  
So the 6
th
 triangular number = 6 × (6 + 1)/ 2 = 21 
 
2. (i) Can a rectangular number also be a square number? 
(ii) Can a triangular number also be a square number? 
Solution: 
 
(i) Yes. A rectangular number can also be a square number. 
Example – 16 is a rectangular number which can also be a square number. 
 
 
(ii) Yes. A triangular number can also be a square number. 
Example – 1 is a triangular number which can also be a square number. 
 
3. Write the first four products of two numbers with difference 4 starting from in the following order: 
1, 2, 3, 4, 5, 6, ……… 
Identify the pattern in the products and write the next three products. 
Solution: 
 
We know that 
1 × 5 = 5  
2 × 6 = 12 
3 × 7 = 21 
4 × 8 = 32 
So the first four products of two numbers with difference 4 
5 – 1 = 4 
6 – 2 = 4 
7 – 3 = 4 
8 – 4 = 4 
 
4. Observe the pattern in the following and fill in the blanks: 
9 × 9 + 7 = 88 
Page 2


 
 
 
 
 
 
Exercise 4.5                                                                           page: 4.23 
1. Without drawing a diagram, find 
(i) 10
th
 square number 
(ii) 6
th
 triangular number 
Solution: 
 
(i) 10
th
 square number 
The square number can be remembered using the following rule 
Nth square number = n × n 
So the 10
th
 square number = 10 × 10 = 100 
 
(ii) 6
th
 triangular number 
The triangular number can be remembered using the following rule 
Nth triangular number = n × (n + 1)/ 2  
So the 6
th
 triangular number = 6 × (6 + 1)/ 2 = 21 
 
2. (i) Can a rectangular number also be a square number? 
(ii) Can a triangular number also be a square number? 
Solution: 
 
(i) Yes. A rectangular number can also be a square number. 
Example – 16 is a rectangular number which can also be a square number. 
 
 
(ii) Yes. A triangular number can also be a square number. 
Example – 1 is a triangular number which can also be a square number. 
 
3. Write the first four products of two numbers with difference 4 starting from in the following order: 
1, 2, 3, 4, 5, 6, ……… 
Identify the pattern in the products and write the next three products. 
Solution: 
 
We know that 
1 × 5 = 5  
2 × 6 = 12 
3 × 7 = 21 
4 × 8 = 32 
So the first four products of two numbers with difference 4 
5 – 1 = 4 
6 – 2 = 4 
7 – 3 = 4 
8 – 4 = 4 
 
4. Observe the pattern in the following and fill in the blanks: 
9 × 9 + 7 = 88 
 
 
 
 
 
 
98 × 9 + 6 = 888 
987 × 9 + 5 = 8888 
9876 × 9 + 4 = ……… 
98765 × 9 + 3 = ……… 
987654 × 9 + 2 = ………. 
9876543 × 9 + 1 = ………. 
Solution: 
 
9 × 9 + 7 = 88 
 
98 × 9 + 6 = 888 
 
987 × 9 + 5 = 8888 
 
9876 × 9 + 4 = 88888 
 
98765 × 9 + 3 = 888888 
 
987654 × 9 + 2 = 8888888 
 
9876543 × 9 + 1 = 88888888 
 
5. Observe the following pattern and extend it to three more steps: 
6 × 2 – 5 = 7 
7 × 3 – 12 = 9 
8 × 4 – 21 = 11 
9 × 5 – 32 = 13 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
Solution: 
 
6 × 2 – 5 = 7 
 
7 × 3 – 12 = 9 
 
8 × 4 – 21 = 11 
 
9 × 5 – 32 = 13 
 
10 × 6 - 45 = 15 
 
11 × 7 - 60 = 17 
 
12 × 8 - 77 = 19 
 
6. Study the following pattern:  
1 + 3 = 2 × 2 
1 + 3 + 5 = 3 × 3 
1 + 3 + 5 + 7 = 4 × 4 
Page 3


 
 
 
 
 
 
Exercise 4.5                                                                           page: 4.23 
1. Without drawing a diagram, find 
(i) 10
th
 square number 
(ii) 6
th
 triangular number 
Solution: 
 
(i) 10
th
 square number 
The square number can be remembered using the following rule 
Nth square number = n × n 
So the 10
th
 square number = 10 × 10 = 100 
 
(ii) 6
th
 triangular number 
The triangular number can be remembered using the following rule 
Nth triangular number = n × (n + 1)/ 2  
So the 6
th
 triangular number = 6 × (6 + 1)/ 2 = 21 
 
2. (i) Can a rectangular number also be a square number? 
(ii) Can a triangular number also be a square number? 
Solution: 
 
(i) Yes. A rectangular number can also be a square number. 
Example – 16 is a rectangular number which can also be a square number. 
 
 
(ii) Yes. A triangular number can also be a square number. 
Example – 1 is a triangular number which can also be a square number. 
 
3. Write the first four products of two numbers with difference 4 starting from in the following order: 
1, 2, 3, 4, 5, 6, ……… 
Identify the pattern in the products and write the next three products. 
Solution: 
 
We know that 
1 × 5 = 5  
2 × 6 = 12 
3 × 7 = 21 
4 × 8 = 32 
So the first four products of two numbers with difference 4 
5 – 1 = 4 
6 – 2 = 4 
7 – 3 = 4 
8 – 4 = 4 
 
4. Observe the pattern in the following and fill in the blanks: 
9 × 9 + 7 = 88 
 
 
 
 
 
 
98 × 9 + 6 = 888 
987 × 9 + 5 = 8888 
9876 × 9 + 4 = ……… 
98765 × 9 + 3 = ……… 
987654 × 9 + 2 = ………. 
9876543 × 9 + 1 = ………. 
Solution: 
 
9 × 9 + 7 = 88 
 
98 × 9 + 6 = 888 
 
987 × 9 + 5 = 8888 
 
9876 × 9 + 4 = 88888 
 
98765 × 9 + 3 = 888888 
 
987654 × 9 + 2 = 8888888 
 
9876543 × 9 + 1 = 88888888 
 
5. Observe the following pattern and extend it to three more steps: 
6 × 2 – 5 = 7 
7 × 3 – 12 = 9 
8 × 4 – 21 = 11 
9 × 5 – 32 = 13 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
Solution: 
 
6 × 2 – 5 = 7 
 
7 × 3 – 12 = 9 
 
8 × 4 – 21 = 11 
 
9 × 5 – 32 = 13 
 
10 × 6 - 45 = 15 
 
11 × 7 - 60 = 17 
 
12 × 8 - 77 = 19 
 
6. Study the following pattern:  
1 + 3 = 2 × 2 
1 + 3 + 5 = 3 × 3 
1 + 3 + 5 + 7 = 4 × 4 
 
 
 
 
 
 
1 + 3 + 5 + 7 + 9 = 5 × 5 
By observing the above pattern, find 
(i) 1 + 3 + 5 + 7 + 9 + 11 
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 
(iii) 21 + 23 + 25 + ….. + 51 
Solution: 
 
(i) 1 + 3 + 5 + 7 + 9 + 11 
By using the pattern 
1 + 3 + 5 + 7 + 9 + 11 = 6 × 6 
                                    = 36 
 
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 
By using the pattern 
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8 
                                                     = 64 
 
(iii) 21 + 23 + 25 + ….. + 51 
We know that 
21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19) 
By using the pattern 
(1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676 
(1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100 
So we get 
21 + 23 + 25 + ….. + 51 = 676 – 100 = 576 
 
7. Study the following pattern: 
1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6 
1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6 
By observing the above pattern, write next two steps. 
Solution: 
 
By using the pattern 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5  
On further calculation 
= (5 × 6 × 11)/6 
So we get 
= 55 
 
By using the pattern 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 
On further calculation 
= (6 × 7 × 13)/6 
So we get 
= 91 
 
8. Study the following pattern: 
1 = (1 × 2)/ 2 
1 + 2 = (2 × 3)/ 2 
Page 4


 
 
 
 
 
 
Exercise 4.5                                                                           page: 4.23 
1. Without drawing a diagram, find 
(i) 10
th
 square number 
(ii) 6
th
 triangular number 
Solution: 
 
(i) 10
th
 square number 
The square number can be remembered using the following rule 
Nth square number = n × n 
So the 10
th
 square number = 10 × 10 = 100 
 
(ii) 6
th
 triangular number 
The triangular number can be remembered using the following rule 
Nth triangular number = n × (n + 1)/ 2  
So the 6
th
 triangular number = 6 × (6 + 1)/ 2 = 21 
 
2. (i) Can a rectangular number also be a square number? 
(ii) Can a triangular number also be a square number? 
Solution: 
 
(i) Yes. A rectangular number can also be a square number. 
Example – 16 is a rectangular number which can also be a square number. 
 
 
(ii) Yes. A triangular number can also be a square number. 
Example – 1 is a triangular number which can also be a square number. 
 
3. Write the first four products of two numbers with difference 4 starting from in the following order: 
1, 2, 3, 4, 5, 6, ……… 
Identify the pattern in the products and write the next three products. 
Solution: 
 
We know that 
1 × 5 = 5  
2 × 6 = 12 
3 × 7 = 21 
4 × 8 = 32 
So the first four products of two numbers with difference 4 
5 – 1 = 4 
6 – 2 = 4 
7 – 3 = 4 
8 – 4 = 4 
 
4. Observe the pattern in the following and fill in the blanks: 
9 × 9 + 7 = 88 
 
 
 
 
 
 
98 × 9 + 6 = 888 
987 × 9 + 5 = 8888 
9876 × 9 + 4 = ……… 
98765 × 9 + 3 = ……… 
987654 × 9 + 2 = ………. 
9876543 × 9 + 1 = ………. 
Solution: 
 
9 × 9 + 7 = 88 
 
98 × 9 + 6 = 888 
 
987 × 9 + 5 = 8888 
 
9876 × 9 + 4 = 88888 
 
98765 × 9 + 3 = 888888 
 
987654 × 9 + 2 = 8888888 
 
9876543 × 9 + 1 = 88888888 
 
5. Observe the following pattern and extend it to three more steps: 
6 × 2 – 5 = 7 
7 × 3 – 12 = 9 
8 × 4 – 21 = 11 
9 × 5 – 32 = 13 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
….. × ……. - ….. = ……. 
Solution: 
 
6 × 2 – 5 = 7 
 
7 × 3 – 12 = 9 
 
8 × 4 – 21 = 11 
 
9 × 5 – 32 = 13 
 
10 × 6 - 45 = 15 
 
11 × 7 - 60 = 17 
 
12 × 8 - 77 = 19 
 
6. Study the following pattern:  
1 + 3 = 2 × 2 
1 + 3 + 5 = 3 × 3 
1 + 3 + 5 + 7 = 4 × 4 
 
 
 
 
 
 
1 + 3 + 5 + 7 + 9 = 5 × 5 
By observing the above pattern, find 
(i) 1 + 3 + 5 + 7 + 9 + 11 
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 
(iii) 21 + 23 + 25 + ….. + 51 
Solution: 
 
(i) 1 + 3 + 5 + 7 + 9 + 11 
By using the pattern 
1 + 3 + 5 + 7 + 9 + 11 = 6 × 6 
                                    = 36 
 
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 
By using the pattern 
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8 
                                                     = 64 
 
(iii) 21 + 23 + 25 + ….. + 51 
We know that 
21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19) 
By using the pattern 
(1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676 
(1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100 
So we get 
21 + 23 + 25 + ….. + 51 = 676 – 100 = 576 
 
7. Study the following pattern: 
1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6 
1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6 
By observing the above pattern, write next two steps. 
Solution: 
 
By using the pattern 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5  
On further calculation 
= (5 × 6 × 11)/6 
So we get 
= 55 
 
By using the pattern 
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 
On further calculation 
= (6 × 7 × 13)/6 
So we get 
= 91 
 
8. Study the following pattern: 
1 = (1 × 2)/ 2 
1 + 2 = (2 × 3)/ 2 
 
 
 
 
 
 
1 + 2 + 3 = (3 × 4)/ 2 
1 + 2 + 3 + 4 = (4 × 5)/ 2 
By observing the above pattern, find 
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 
(ii) 50 + 51 + 52 + ……. + 100 
(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100 
Solution: 
 
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 
We get 
= (10 × 11)/2 
On further calculation 
= 55 
 
(ii) 50 + 51 + 52 + ……. + 100 
We can write it as 
(1 + 2 + 3 + …… + 99 + 100) – (1 + 2 + 3 + 4 + ….. + 47 + 49) 
So we get 
(1 + 2 + 3 + …… + 99 + 100) = (100 × 101)/2 
(1 + 2 + 3 + 4 + ….. + 47 + 49) = (49 × 50)/2 
By substituting the values 
50 + 51 + 52 + ……. + 100 = (100 × 101)/2 + (49 × 50)/2 
On further calculation 
= 5050 – 1225 
We get 
= 3825 
 
(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100 
We can write it as 
2 (1 + 2 + 3 + 4 + …… + 49 + 50) 
So we get 
= 2 × (50 × 51)/2 
On further calculation 
= 2 × (1275)  
We get 
= 2550 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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